Fractions continued: Addition and subtraction of fractions

Subject : 

Mathematics

Term :

First Term

Week:

Week  8

Class :

Jss 1

Previous lesson : 

The pupils have previous knowledge of  Review of the first half term’s work and periodic test

Topic : 

Fractions continued: Addition and subtraction of fractions

Behavioural objectives :

At the end of the lesson, the pupils should be able to

• add up fractions that are of the same denominators
• sum up fractions that are of different denominators
• increase a fraction by another fraction
• calculate the difference between fractions
• take away a certain fraction from another one
• solve word problem questions on fractions

Instructional Materials :

• Wall charts
• Pictures
• Related Online Video
• Flash Cards

Methods of Teaching :

• Class Discussion
• Group Discussion
• Asking Questions
• Explanation
• Role Modelling
• Role Delegation

Reference Materials :

• Scheme of Work
• Online Information
• Textbooks
• Workbooks
• 9 Year Basic Education Curriculum
• Workbooks

Content :

WEEK EIGHT

TOPIC : ADDITION AND SUBTRACTION OF FRACTIONS

CONTENT

i. Introduction

ii. Addition of Fractions

iii. Subtraction of Fractions

iv. Further Examples.

I. Introduction

Two or more fractions can be added or subtracted immediately if they both possess the same denominator, in which case we add or subtract the numerators and divide by the common denominator . For example

2/5 + 1/5 = 2 + 1 = 3/5

5

If they do not have the same denominator they must be rewritten in equivalent form so that they do have the same denominator – called the common denominator e.g

2/7 + 1/5 = 10/35 + 7/35 = 10 + 7 = 17

35 35.

The common denominator of the equivalent fraction is the LCM of the two original denominator that is,

2/7 + 1/5 = 5 x 2 + 7 x 1 = 10 + 7 = 10 + 7 = 17

5 x 7 7 x 5 35 35 35 35

From the explanation, the above example has its LCM = 35.

Can you try this,

5/8 + 1/6 ?

the correct answer is 19/24

Summary

If fractions have different denominators:

1. Find a common denominator by expressing each fractions as an equivalent fraction
2. Add or subtract their numerators.

II. Addition of Fractions

Example: Simplify the following fractions

(a) ¼ + ½ (b) 2/3 + 5/6 (c) 2/5 + ½ + ¼

Solution

a. ¼ + ½ = ¼ + 2 x 1 = ¼ + 2 = 1+ 2 = 3

2 x 2 4 4 4

(b) 2 + 5 = 2 x 2 + 5 = 4 + 5 = 4 + 5 = 9 = 1 ³/6

3 6 3 x 2 6 6 6 6 6

= 1 ½ mixed fraction

(c ) 2 + ½ + ¼ = 2 x 4 + 1 x 10 + 1 x 5

5 5 x 4 2 x 10 4 x 5

= 8/20 |10/20 + 5/20 = 8 + 10 + 5 = 23/24 = 1 3/20

20.

Example 2:Simplify the following fractions.

1. 1 ¾ + 2 ²/3 + ½
2. 3 ¾ + ⁵/8 1 ⁷/12
3. 5 ⁴/9 +7 ¹/3 + ¹/12

Solution.

1 ¾ + 2 ²/3 + ½

convert to improper fractions

⁷/4 +⁸/3 + ½

7 x 3 + 8 x 4 + 1 x 6

4 x 3 3x 4 2 x 6

²¹/12 + ³²/12 + ⁶/12

= 21+ 32 + 6

12

= ⁵⁹/ 12

4 ¹¹/12

b. 3 ¾ + ⁵/8 + 1 ⁷/12

convert to improper fractions

¹⁵/4 + ⁵/ 8 +¹⁹/ 12

15 x 6 + 5 x 3 + 19 x 2

4 x 6 8 x 3 12 x 12

= ⁹⁰/24 +¹⁵/ 24 + ³⁸/24

= 90 + 15 + 38

24

143

24

5 ²³/24

c. 5 ⁴/9 +7 ¹/3 + ¹/12

convert to improper fractions

⁴⁹/9 + ²²/2 + ¹/12

= 49 x 4 + 22 x 12 + 1 x 3

9 x 4 3 x 12 12 x 3

¹⁹⁶/36 +²⁶⁴/36 + ³/36

196 + 264 + 3

36

463

36

= 12

EVALUATION

Simplify the following:

a. 3 ⁷/8 + 2 ³/4

b. 1 ½ + 2 ¹/3 + 3 ¼

c. 5 + 1 ¾ + 2 ²/3

READING ASSIGNMENT

1. Essential Mathematics for JSS 1 by AJS Oluwasanmipg 32 – 45

2. New General Mathematics for JSS1 by M.F. Macraepg 32 – 33.

III. Subtraction of Fractions

Example 1: simplify the following:

a. ²/3 – ¼ b. ¾ – ⁵/8 c. 5 ¾ – 2 ⁴/5

Solution

²/3 – ¼

= 2 x 4 – 1 x 3

3 x 4 4 x 3

= 8 – 3

12 12

8 – 3

12

5

12.

b. ¾ – ⁵/8

3 x 2 – 5

4 x 2 8

6 – 5

8 8

6 – 5

8

1

8

c. 5 ¾ – 2 ⁴/5

convert to improper fraction,

23/4 –¹⁴/5 = 23 x 5 – 14 x 4

4 x 5 5 x 4

= 115 -56

20 20

= 115 – 56

20

59

20 = 2 ¹⁹/20.

Example 2: simplify the following :

a. 5 ¹/₆ – 3 ²/₃ + 6 ⁷/₁₂

b. 2 ½ + 3 + ⁷/₁₀ – ²/₅ – 2

c. 2 ½ + ¾ – ¹¹/₆ + 4 – 1 ²/₃

Solution

a. 5¹/₆ – 3 ²/₃ + ^{6 7}/₁₂

= ³¹/₆ –¹¹/₃ + ⁷⁹/₁₂

= 31 x 2 – 11 x 14 + 79

6 x 2 3 x 4 12

⁶²/₁₂ –⁴⁴/₁₂ + ⁷⁹/₁₂

= 62 – 44 + 79

12

97

12 = 8 ¹/₁₂

b. 2 ½ + 3 + ⁷/₁₀ – ²/₅ – 2

= ⁵/₂ – ³/₁ +⁷/₁₀ – ²/₅ – ²/₁

5 x 5 – 3 x 10 + 7 – 2 x 2 – 2 x 10

2 x 5 1 x10 10 5 x 2 1x 10

25- 30+ 7 – 4 – 20

10 10 10 10 10

= 25 – 30 + 7 – 4 – 20

10

= 25 + 7 – 30 – 4 -20

10

= 32 – 30 – 4 – 20

10

= -22

10

2 ²/₁₀ = – 2 ¹/₅

c. 2 ½ + ¾ – 1 ¹/₆ + 4 – 1 ²/₃

⁵/₂ + ¾ –⁷/₆ +⁴/1 – ⁵/₃

5 x 6 +3x 3 – 7 x 2 + 4 x 12 – 5 x 4

2 x 6 4 x 3 6 x 2 1×12 3 x 4

30 + 9 – 14 + 48 – 20

12 12 12 12 12

30+99 – 14 + 48 – 20

12

30 + 9- 14 + 48 – 20

12

30 +9 + 48 – 14 – 20

12

87 – 34

12

⁵³/12

4 ⁵/₁₂

EVALUATION

Simplify the following :

1. 2 ½ – 1 ⁴/₅ + 2 ³/2 – 1

2.7 ½ + 3 ¹/₆ – 3 ¼

3.14 ⁴/₁₅ – 4 ²/₃ + 7 ¹/₅

III. Further examples

Example 1

What is the sum of 2 ¾ and 2 ⁴/₅?

Solution

Sum = addition 9 + 0

Hence, sum of 2 ¾ and 2 ⁴/₅ is

= 2 ¾ + 2 ⁴₅

11+ 14

4 5

11 x 5 + 14 x 4

4 x 5 5 x 4

= 55 + 56

20 20

55 + 56

20

111

20 = 5 11/20

Example 2

A 2 ¼ kg piece of meat is cut from a meat that weighs 3 2/5kg. What is the weight of the meat left?

Solution

Original weights of meat = 2 2/5kg

Weight of meat cut = 2 ¼ kg

Final weight of meat = 3 2/5 – 2 ¼

= 17/5 – 9/4

= 17 x 4 – 9 x 5

5 x 4 4 x 5

68 – 45

20 20.

68 – 45

20

23

• 1. = 2 3/20

The weight of the meat left = 2 3/20 kg.

Example 3

A fruit grower uses 1/3 of his land for bananas, 3/8 for pineapples, 1/6 for mangoes and the remainder for oranges. What fraction of his land is used for oranges.

Solution.

The entire land is a unit = 1

Every other fractions add up to give 1

;.oranges + bananas + pineapple + mango = 1

:. Orange = 1 – ( 1/3 + 3/8 + 1/6)

= 1 – ( 1 x 8 + 3 x 3 + 1 x 4 )

3 x 8 8 x 3 6 x 4

= 1 – ( 8/4 + 9/24 + 4/24 )

= 1 – 8 (8 + 9 + 4 )

24

1/1 – 21/24

= 24 – 21

24

= 3/24 = 1/8.

:. The fruit grower used 1/8 for oranges.

EVALUATION

1. By how much is the sum of 2 4/5 and 4 ½ less than 8 1/10?

2. A boy plays football for 13/4 hours, listens to radio for ¾ hours and then spends 1 ¼ hours doing his homework. How much time does he spend altogether doing these things?

READING ASSIGNMENT

1. Essential Mathematics for JSS 1 by AJS Oluwasanmipg 45

2. New General Mathematics for JSS1 by M.F. Macraepg 33

WEEKEND ASSIGNMENT

1. Simplify 2 ½ + ¼

(a) 3 ¾ (b).2 1/8 (c) 1 ¾ (d) 2 ¾.

2. Simplify 4 2/5 – 3 ¼

(a) 1 3/20 (b) 3 2/5 (c) 1 7/20 (d) 1 5/8

3. The common denominator of the fractions is

(a) 8 (b) 12 (c ) 6 (d) 15

4. Simplify

(a) 1 43/45 (b) 43/45 (c) 2 37/45 1 41/45

5. What is the sum of 1 ¾, 2 3/5 and 5 ¾

(a0 3 1/30 (b) 5 1/60 (c) 7 1/60 (d) 8 1/50.

THEORY

1. Simplify the following;

( a) 3⁷/8 + 2 ¾ (b) 2 ⁵6 + 5 ⁷/8

(c)

2. Mr. Hope spends 1/3 of his earnings on food and ¼ on clothes. He then saves the rest. What fraction does he

(a) spend altogether

(b) save?

Presentation

The topic is presented step by step

Step 1:

The class teacher revises the previous topics

Step 2.

He introduces the new topic

Step 3:

The class teacher allows the pupils to give their own examples and he corrects them when the needs arise

Conclusion

The class teacher wraps up or conclude the lesson by giving out short note to summarize the topic that he or she has just taught.

The class teacher also goes round to make sure that the notes are well copied or well written by the pupils.

He or she does the necessary corrections when and where  the needs arise.